# RMS & Avg. of waves

## Homework Statement

I need to find the average and RMS voltages for the waves (See attached pictures) but I'm not sure exactly how to do it as I don't know how to get an equation for V(t)

## Homework Equations

I Know:

VRMS = $$\sqrt{\frac{1}{T}\int V(t)^2 dt}$$ between 0 and T

And for Avg it's just VAVG = $$\frac{1}{T}\int V(t) dt$$ between 0 and T

## The Attempt at a Solution

I thought I could find the averages by finding the total area for 1 period, then dividing by 1 period. This works (I think, as I don't have the answers)

Giving for the Triangular wave: $$\frac{T}{3} \times \frac{10}{T} = 3.3V$$

and for the second sawtooth type wave: $$\frac{\frac{1}{2}\frac{T}{3} \times 1}{T} = \frac{1}{6}V$$

Thanks

#### Attachments

• 5.3 KB Views: 241
• 3.8 KB Views: 283

Related Engineering and Comp Sci Homework Help News on Phys.org
Redbelly98
Staff Emeritus
Homework Helper
Looks good for VAVG.

For VRMS, you'll need to write equations for the waveforms. To do that, you may choose any point to represent t=0.

That's the problem, I don't know what to write the equations as. The triangular wave has 3 functions. One for 0 - $$\frac{T}{3}$$ (upwards slanting triangle) one for 0 - $$\frac{2T}{3}$$ (downwards slanting triangle) and one for the last part of the waveform, which is just V(t) = 0.

What are the waveforms for the first 2 sections? And can they be summed like so?:

VRMS = $$\sqrt{\frac{1}{T}\int V_A(t)^2 dt} + \sqrt{\frac{1}{T}\int V_B(t)^2 dt}$$ the first integral being from 0 to T/3 and the second from T/3 to 2T/3

Redbelly98
Staff Emeritus
Homework Helper
That's pretty much the idea, except that it would be

$$\sqrt{\frac{1}{T}\int V_A(t)^2 dt \ + \ \frac{1}{T}\int V_B(t)^2 dt}$$

with, as you said, "the first integral being from 0 to T/3 and the second from T/3 to 2T/3"

I still can't seem to figure out the RMS values. The problem I have is determining the functions V(t) for each section of the wave.

I have an exam on this tomorrow so any help would be appreciated!

Redbelly98
Staff Emeritus
Homework Helper
Note that the function is composed of straight line segments. Do you remember, from high school algebra,

y = mx + b

where m is the slope and b is the y-intercept?

For example . . .

. . . for the triangular wave, and 0 ≤ t ≤ T/3:

v(t) = m t + b

What is the slope m in this interval (0 ≤ t ≤ T/3)?
What is the intercept b in this interval?

Find m and b in this interval, and you'll have the function.

Do the same for the interval T/3 ≤ t ≤ 2T/3, and you'll have v(t) in that interval as well.

Hmm...
using v(t) = m t + b
I get:
$$V_A(t) = \frac{30t}{T}$$
$$V_B(t) = 20 - \frac{30t}{T}$$

So I tried integrating all of this, then taking the root (long and tiring process!) and I got V_RMS = 4.71V

Here's the integration I did (without the square root)

$$\sqrt{\frac{30^2}{T^3}\int t^2 dt + \frac{1}{T}\int 400t dt - \frac{1}{T^2}\int 600t^2 dt + \frac{1}{T^3}\int 300t^3 dt}$$

With the first being between 0 and T/3 and the last 3 parts being between T/3 and 2T/3
(there's 3 parts because of squaring $$20 - \frac{30t}{T}$$)

I hope this is right!

Redbelly98
Staff Emeritus